y=(2x^2+4x+1)^12,dy/dx=?

(x^2)dy+（y^2)dx=dx-dy dy/dx-y/x=x^2 dy/dx,y=(1+x+x^2)e^x 微分方程 dy/dx=(-2x)/y dy/dx-2y/(x+1)=(x+1)^2 dy/dx-2y/(1+x)=(x+1)^3 dy/dx=(e^x+x)(1+y^2)通解 微分方程dx/2(x+y^4)=dy/y dy/dx=x(1+y^2)/y通解 y=[sin(x^4)]^2,则dy/dx=?,dy^2/dx^2=?,dy/d(x^2)=? y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy y=(2x^2+4x+1)^12,dy/dx=?